Problem: Simplify and expand the following expression: $ \dfrac{1}{2t + 2}- \dfrac{1}{t + 9}- \dfrac{2t}{t^2 + 10t + 9} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2t + 2} = \dfrac{1}{2(t + 1)}$ We can factor the quadratic in the third term: $ \dfrac{2t}{t^2 + 10t + 9} = \dfrac{2t}{(t + 1)(t + 9)}$ Now we have: $ \dfrac{1}{2(t + 1)}- \dfrac{1}{t + 9}- \dfrac{2t}{(t + 1)(t + 9)} $ The least common multiple of the denominators is: $ 2(t + 1)(t + 9)$ In order to get the first term over $2(t + 1)(t + 9)$ , multiply by $\dfrac{t + 9}{t + 9}$ $ \dfrac{1}{2(t + 1)} \times \dfrac{t + 9}{t + 9} = \dfrac{t + 9}{2(t + 1)(t + 9)} $ In order to get the second term over $2(t + 1)(t + 9)$ , multiply by $\dfrac{2(t + 1)}{2(t + 1)}$ $ \dfrac{1}{t + 9} \times \dfrac{2(t + 1)}{2(t + 1)} = \dfrac{2(t + 1)}{2(t + 1)(t + 9)} $ In order to get the third term over $2(t + 1)(t + 9)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{2t}{(t + 1)(t + 9)} \times \dfrac{2}{2} = \dfrac{4t}{2(t + 1)(t + 9)} $ Now we have: $ \dfrac{t + 9}{2(t + 1)(t + 9)} - \dfrac{2(t + 1)}{2(t + 1)(t + 9)} - \dfrac{4t}{2(t + 1)(t + 9)} $ $ = \dfrac{ t + 9 - 2(t + 1) - 4t} {2(t + 1)(t + 9)} $ Expand: $ = \dfrac{t + 9 - 2t - 2 - 4t}{2t^2 + 20t + 18} $ $ = \dfrac{-5t + 7}{2t^2 + 20t + 18}$